Problem: Solve the equation. $\dfrac{dy}{dx}=\dfrac{x}{7\cos(y)}$ Choose 1 answer: Choose 1 answer: (Choice A) A $y=\dfrac{\cos(x)+x\sin(x)}{7\cos^2(x)}+C$ (Choice B) B $y=\arcsin\left(\dfrac{x^2}{14}+C\right)$ (Choice C) C $y=\dfrac{\cos(x)+x\sin(x)+C}{7\cos^2(x)}$ (Choice D) D $y=\arcsin\left(\dfrac{x^2}{14}\right)+C$
We can bring this equation to the form $f(y)\,dy=g(x)\,dx$ : $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{x}{7\cos(y)} \\\\ 7\cos(y)\,dy&=x\,dx \end{aligned}$ This means we can solve this equation using separation of variables! $\begin{aligned} 7\cos(y)\,dy&=x\,dx \\\\ \int 7\cos(y)\,dy&=\int x\,dx \\\\ 7\sin(y)&=\dfrac{x^2}{2}+C_1 \\\\ \sin(y)&=\dfrac{x^2}{14}+C \\\\ \arcsin(\sin(y))&=\arcsin\left(\dfrac{x^2}{14}+C\right) \\\\ y&=\arcsin\left(\dfrac{x^2}{14}+C\right) \end{aligned}$ [Where did we get C?] Notice that after the integration, more work was required in order to isolate $y$. In conclusion, this is the solution of the equation: $y=\arcsin\left(\dfrac{x^2}{14}+C\right)$